if a+bi lies on a cirle in complex plane then is it necessary that i(a+bi) lies on the circle?

no i dont think so

there are infinite circles that can be constructed through a point

but if this is a circle with origin as the center

then the distance of any point on the circle will be radius r = root (a^2 + b^2)

all points at radius r will lie on the circle

lets see i(a+bi) = -b + ai

the distance from rigin is root of (-b)^2 + a^2

so its located at the same distance fron origin so it will lie on the same circle as a + bi

if the center is the origin

all the best

The answer is yes (if you mean circles centered at the origin)

One way to see this is to recognize that multiplication by i simply rotates a complex number by positive pi/2 radians, without changing the modulus (radius)

Another way to see this is simply to compute. the number a + bi lies on a circle whose radius is sqrt ( a^2 + b^2). The number i(a + bi) = -b + ai lies on a circle whose radius is sqrt((-b)2 + a^2) which is the same as sqrt ( a^2 + b^2).

A third way is to use the property |zw| = |z| |w| for any two complex numbers z and w. So |i(a+bi)| = |i| |a+bi| = |a+bi|, thus the two numbers have the same modulus (radius).

> (a+bi) : implies (a,b) lies on circle... (1)

>i(a+bi) = (-b+ai) : implies (-b,a).........(2)

>x^2+y^2 = r^2 : [EQUATION OF CIRCLE] . (3)

condition(1) and (3)says : a^2+b^2 = r^2

> it also implies: (-b)^2 + a^2 = r^2 ..........(4)

>condition(4) implies (-b,a) = i(a+bi) always lie on the given circle.